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2 years ago

Three resistors of 3.3 kΩ, 4.7 kΩ and 3.9 kΩ are connected in series across a

Physics

1 answer:

natima [27]2 years ago

6 0

Explanation:

A.) equivalent resistance

Equivalent resistance ( Rt ) in a series connection is sum of all resistance.

Rt = 3300 + 4700 + 3900

= 11900 ohms

= 11.9 k ohm

B.) current is constant across the whole circuit if it is in series .

current = potential difference ÷ total resistance

current = 12 / 11900 = 0.00100 Amphere

= 0.1 mA

C. potential difference ( v ) across the circuit =

current( i ) × resistance ( Rt )

=》v = 0.1 × 0.001 × 11900 = 12 volts

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3 years ago

What is the soil fabric?

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2 years ago

The winning time for a 500.0 mile circular race track was 3 hours and

lyudmila [28]

Explanation:

We have,

Distance traveled in a circular track is 500 miles

The winning time was 3 hours and 13 minutes. It means time is 3.217 hours.

The driver's average speed is given by total distance divided by total time taken. Its formula can be written as :

$v=\dfrac{d}{t}\\\\v=\dfrac{500\ miles}{3.217\ h}\\\\v=155.42\ mph$

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3 years ago

Master of physics needed

Delicious77 [7]

Hey JayDilla, I get 1/3. Here's how:
Kinetic energy due to linear motion is:
$E_{linear}= \frac{1}{2}mv^2$
where
$v=r \omega$
giving
$E_{linear}= \frac{1}{2}mr^2 \omega ^2$

The rotational part requires the moment of inertia of a solid cylinder
$I_{cyl} = \frac{1}{2}mr^2$
Then the rotational kinetic energy is
$E_{rot}= \frac{1}{2}I \omega ^2= \frac{1}{4}mr^2 \omega ^2$
Adding the two types of energy and factoring out common terms gives
$\frac{1}{2}mr^2 \omega ^2(1+ \frac{1}{2})$
Here the "1" in the parenthesis is due to linear motion and the "1/2" is due to the rotational part. Since this gives a total of 3/2 altogether, and the rotational part is due to a third of this (1/2), I say it's 1/3.

8 0

3 years ago

A 15.0 Ohms resistor is connected in series to a 120V generator and two 10.0 Ohms resistors that are connected in parallel to ea

Nostrana [21]

Hi there! :)

Reference the diagram below for clarification.

We must begin by knowing the following rules for resistors in series and parallel.

In series:
$R_T = R_1 + R_2 + ... + R_n$

In parallel:
$\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + ... + \frac{1}{R_n}$

We can begin solving for the equivalent resistance of the two resistors in parallel using the parallel rules.

$\frac{1}{R_{T, parallel}} = \frac{1}{10} + \frac{1}{10}\\\\\frac{1}{R_{T, parallel}} = \frac{2}{10} = \frac{1}{5}\\\\R_{T, parallel} = 5\Omega$

Now that we have reduced the parallel resistors to a 'single' resistor, we can add their equivalent resistance with the other resistor in parallel (15 Ohm) using series rules:
$R_T = 15 + 5\\\\\boxed{R_T = 20 \Omega}$

We can use Ohm's law to solve for the current in the circuit.

$i = \frac{V}{R_T}\\\\i = \frac{120}{20} = \boxed{6 A}$

For resistors in series, both resistors receive the SAME current.

Therefore, the 15Ω resistor receives 6A, and the parallel COMBO (not each individual resistor, but the 5Ω equivalent when combined) receives 6A.

In this instance, since both of the resistors in parallel are equal, the current is SPLIT EQUALLY between the two. (Current in parallel ADDS UP). Therefore, an even split between 2 resistors of 6 A is <u>3A for each 10Ω resistor</u>.

Since the 15.0 Ω resistor receives 6A, we can use Ohm's Law to solve for voltage.

$V = iR\\\\V = (6)(15) = \boxed{90 V}$

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1 year ago